Page III-10-1 / Chapter Ten Lecture Notes …mhchem.org/222/pdfPPT/Ch13BWeb.pdfIntermolecular...

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Chemistry 222 Professor Michael Russell

Intermolecular Forces, Liquidsand SolidsChap. 10

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States of MatterThe fundamental difference between states of matter is the distance between particles.

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States of MatterBecause in the solid and liquid states particles are closer together, we refer to them as condensed phases.

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The States of Matter

The state a substance is in at a particular temperature and pressure depends on two antagonistic entities:

• the kinetic energy of theparticles;

• the strength of theattractions between theparticles.

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Intermolecular Forces

The attractions between molecules (intermolecular forces) are not nearly as strong as the intramolecular attractions that hold compounds together.

Intramolecular forces: ionic, covalent, metallic Intermolecular forces are not chemical bonds!

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EK = lowest intermediate highest

Intermolecular (IM) Attraction:highest intermediate lowest

Kinetic Energy (Ek) vs. Attractive (IM) Force

We will assume gases have no IM force in CH 222

Page III-10-1 / Chapter Ten Lecture Notes


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Intermolecular forces are strong enough to affect physical properties such as boiling and melting points, vapor pressures, and viscosities.

See the IM Forces Guide

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Attraction Between Ions and Permanent

Dipoles

Water is highly polar (it has a dipole) and can interact with positive ions to give hydrated ions in water.

HH

water dipole

••

••

O-δ

+δ

This is the Ion-Dipole IM force

MAR This is the Ion-Dipole IM force

Water is highly polar (it has a dipole) and can interact with positive ions to give hydrated ions in water.

HH

water dipole

••

••

O-δ

+δ


Dipoles

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DipolesMany metal ions are

hydrated. This is the reason metal salts dissolve in water.

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Attraction Between Ions and

Permanent Dipoles

Attraction between ions and dipole depends on ion charge and ion-dipole distance.

Measured by ∆H for Mn+ + H2O --> [M(H2O)x]n+

-405 kJ/mol -1922 kJ/mol -263 kJ/mol

OH

Hδ+

δ-• • • O

H

Hδ+

δ-• • • O

H

Hδ+

δ-• • •

Na+ Mg2+Cs+

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Dipole-Dipole Forces

Dipole-dipole forces bind molecules having permanent dipoles to one another.



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Dipole-Dipole Forces

Influence of dipole-dipole forces is seen in the boiling points of simple molecules.

Compd Mol. Wt. Boil Point N2 28 -196 oC

CO 28 -192 oC Br2 160 59 oC

ICl 162 97 oC

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Hydrogen BondingA special form of the dipole-dipole force

which enhances dipole-dipole attractions.

H-bonding is strongest when X and Y are N, O, or F

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Hydrogen Bonding in H2O

Ice has open lattice-like structure. Ice density is < liquid and so solid floats on water.

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Hydrogen BondingH bonds ---> abnormally high boiling point of

water.

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Boiling Points of Simple Hydrocarbon

Compounds

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Hydrogen BondingHydrogen bonding and base pairing in DNA.



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FORCES INVOLVING INDUCED DIPOLES

How can non-polar molecules such as O2 and I2 dissolve in water?

The water dipole INDUCES a dipole in the O2 electric cloud.

Dipole-induced dipole

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Consider I2 dissolving in alcohol, CH3CH2OH.

OH

-δ

+δ

I-I

R

-δ

+δ

OH+

δ

-δ

I-I

R

The alcohol temporarily creates or INDUCES a dipole in I2.

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Formation of a dipole in two nonpolar I2 molecules.

Induced dipole-induced dipole

"Induced Dipole-Induced Dipole" is also known as "London Dispersion", same thing MAR


The induced forces between I2 molecules are very weak, so solid I2 sublimes (goes from a solid to gaseous molecules).

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The magnitude of the induced dipole depends on the tendency to be distorted.

Higher molar mass ---> larger induced dipoles. Larger atoms have larger electron clouds which

are easier to polarize

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Intermolecular Forces Summary



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Strength Summary

Ion-ion / metallic force strongest

of all

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LiquidsIn a liquid • molecules are in

constant motion • there are appreciable

intermolecular forces • molecules close

together • Liquids are almost

incompressible • Liquids do not fill the

container

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LiquidsThe two key properties we need to describe

are EVAPORATION and its opposite-CONDENSATION

LIQUID VAPORbreak IM bonds

make IM bonds

Add energy

Remove energy<---condensation

evaporation--->

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LiquidsTo evaporate,

molecules must have sufficient energy to break IM forces.

Breaking IM forces requires energy. The process of evaporation is endothermic.

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Liquids Distribution of molecular energies in a liquid.

Kinetic Energy proportional to Temperature

At higher T a much larger number of molecules has high enough energy to break IM forces and move from liquid to vapor state.

0Num

ber o

f mol

ecul

es

Molecular energyminimum energy neededto break IM forces and evaporate

higher Tlower T

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LiquidsWhen molecules of liquid

are in the vapor state, they exert a VAPOR PRESSURE

EQUILIBRIUM VAPOR PRESSURE is the pressure exerted by a vapor over a liquid in a closed container when the rate of evaporation = the rate of condensation.



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Vapor Pressure

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Boiling Liquids

Liquid boils when its vapor pressure equals atmospheric pressure.

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Boiling Point at Lower Pressure

When pressure is lowered, the vapor pressure can equal the external pressure at a lower temperature.

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Consequences of Vapor Pressure Changes

When can cools, vp of water drops. Pressure in the can is less than that of atmosphere, so can is crushed.

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Consequences of Vapor Pressure Changes - Whoops!

When car cools on hot day (i.e. cleaning with cold water), vp of fumes inside drops. Pressure in the car is less than that of atmosphere, so car is crushed! MAR

Equilibrium Vapor Pressure



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The curves show all conditions of P and T where LIQ and VAP are in EQUILIBRIUM.

The VP rises with T. When VP = external P, the liquid boils. This means that BPs of liquids change with

altitude.

Liquids

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LiquidsIf external P = 760

mm Hg, T of boiling is the NORMAL BOILING POINT

VP of a given molecule at a given T depends on IM forces. Here the VPs are in the order:

C2H5H5C2 HH5C2 HH

wateralcoholether

increasing strength of IM interactions

extensiveH-bondsH-bonds

dipole-dipole

OOO

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Viscosity

Ethanol

Glycerol

VISCOSITY is the tendency or resistance of liquids to flow.

Liquids "flow" differently due to the strength of their intermolecular bonds

Viscosity results from several factors, including IM interactions, molecular shape and size

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LiquidsMolecules at surface behave differently than

those in the interior.

Molecules at surface experience net INWARD force of attraction.

This leads to SURFACE TENSION - the energy required to break the surface.

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Surface Tension

SURFACE TENSION also leads to spherical liquid droplets. MAR

LiquidsIM forces also lead to CAPILLARY action

Cohesive forces: interactions between like particles.

Adhesive forces: interactions between unlike particles.

Concave Meniscus: adhesive forces ≥ cohesive forces (H2O on glass)

Convex Meniscus: Cohesive forces > adhesive forces (Hg on glass).



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Capillary Action

Movement of water up a piece of paper depends on H-bonds between H2O and the OH groups of the cellulose in the paper.

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Heat Transfer with Phase Change

Chapter 5: Heat Transfer with

no Phase Change (q = mC∆T)

Overall patterns: solid → liquid → gas = endothermic reaction gas → liquid → solid = exothermic reaction

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Heat Transfer with Phase Change

HEAT OF VAPORIZATION is the heat required (at constant P) to vaporize a liquid.

LIQ + heat ---> VAP Compd. ∆Hvap (kJ/mol) IM Force

H2O 40.7 (100 oC) H-bonds

SO2 26.8 (-47 oC) dipoleXe 12.6 (-107 oC) induced dipole

HEAT OF FUSION is the heat required (at constant P) to melt a solid.

SOL + heat ---> LIQ Temperature constant during phase change MAR

Clausius-ClapeyronThe Clausius–Clapeyron Equation

provides a link between vapor pressure (P), temperature (T), and molar heat of vaporization (∆Hvap).

By taking measurements at two temps:

R = 8.3145 J mol-1 K-1

ln P = -ΔHvap

!

RT + C

ln P1

P2

= ΔHvap

!

R1T2

- 1T1

"

#$

%

&'

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Heat water q = mC∆T Evaporate water

Melt ice

Heating/Cooling Curve for Water

Note that T is constant as ice melts

Note that T is constant as liquid water evaporates

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Heat of fusion of ice = 333 J/gSpecific heat of water = 4.184 J/g•K

Heat of vaporization = 2260 J/g

What quantity of heat is required to melt 500. g of ice at 0.0 oC and heat the water to steam at 100. oC?

+333 J/g

Heat & Changes of State

solid (ice) liquid (0 – 100 °C) gas (steam)

+2260 J/g



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What quantity of heat is required to melt 500. g of ice at 0.0 oC and heat the water to steam at 100. oC?

1. To melt ice q = (500. g)(333 J/g) = 1.67 x 105 J2. To raise water from 0.0 oC to 100. oC q = (500. g)(4.184 J/g•K)(100. - 0)K = 2.09 x 105 J3. To evaporate water at 100. oC q = (500. g)(2260 J/g) = 1.13 x 106 J4. Total heat energy = 1.51 x 106 J = 1510 kJ

Heat & Changes of State

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Phase Diagrams

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TRANSITIONS BETWEEN PHASES

See the phase diagram for water in text Lines connect all conditions of T and P

where EQUILIBRIUM exists between the phases on either side of the line.

At equilibrium particles move from liquid to gas as fast as they move from gas to liquid.

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Phase Diagram for Water

Animation of solid phase.

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Animation of equilibrium between solid and liquid phases. MAR


Animation of liquid phase.



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Animation of equilibrium between liquid and gas phases.

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Animation of gas phase.

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Animation of equilibrium between solid and gas phases.

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Animation of triple point.

At the TRIPLE POINT all three phases are in equilibrium.

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Phases Diagrams-

Important Points for Water

T(˚C) P(mm Hg) Normal boil point 100 760 Normal freeze point 0 760 Triple point 0.0098 4.58

Water at the Triple Point MAR

SolidsWe can think of solids as falling into two groups:

• crystalline: particles in highly ordered arrangements

• amorphous: no particular order in arrangement of particles.

We will focus on crystalline solids



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Properties of Crystalline Solids1. Molecules, atoms or

ions locked into a CRYSTAL LATTICE

2. Particles are CLOSE together

3. STRONG IM forces 4. Highly ordered, rigid,

incompressible

ZnS, zinc(II) sulfide

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Crystal LatticesRegular 3-D arrangements of equivalent

LATTICE POINTS in space. Lattice points define UNIT CELLS

smallest repeating internal unit that has the symmetry characteristic of the solid.

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The Brevais Lattices7 main types: • Triclinic • Monoclinic • Orthorhombic • Tetragonal • Hexagonal • Rhombohedral • Cubic (Isometric) We will use just the cubic

system in CH 222

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Cubic Unit CellsThere are 7 basic crystal systems, but we are only concerned with CUBIC.

All anglesare 90 degrees

All sidesequal length

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Cubic Unit CellsSimple cubic (SC)

Body-centered cubic (BCC)

Face-centered cubic (FCC)

See Cubic Unit Cells Guide

Three types of Cubic Unit Cells:

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Unit Cells for Metals



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Simple cubic unit cell. Note that each atom is at a corner of a unit

cell and is shared among 8 unit cells.

Simple Cubic Unit Cell

MARAtom at each corner,

Only 1 net atom per simple cubic cell

The Simple Cubic Unit Cell

MARAtom at each cube corner plus one in center

Two net atoms per bcc unit cell

Body-Centered Cubic Unit Cell

also known as cubic close

packing

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Atom in each cube corner plus atom in each cube face, four net atoms per fcc unit cell

Face-Centered Cubic Unit Cell

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Finding the Lattice TypeTo find out if a metal is SC, BCC or FCC, use the

known radius and density of an atom to calc. no. of atoms per unit cell.

PROBLEM Al has density = 2.699 g/cm3 and Al radius = 143 pm. Verify that Al is FCC.

SOLUTION 1. Calc. unit cell edge (cm)

see handout: edge = 4 * radius / √2

edge = 4 * 143 pm / √2 = 404 pm

404 pm * (10-10 cm / pm) = 4.04 * 10-8 cm

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Finding the Lattice TypePROBLEM Al has density = 2.699 g/cm3 and Al radius = 143 pm.

Verify that Al is FCC. SOLUTION

2. Calc. unit cell volume edge = 4.04 x 10-8 cm (previous slide) V = (cell edge)3 = (4.04 x 10-8 cm)3 V = 6.62 x 10-23 cm3

3. Now use density to find mass mass = (6.62 x 10-23 cm3)(2.699 g/cm3)

= 1.79 x 10-22 g/unit cell



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Finding the Lattice TypePROBLEM Al has density = 2.699 g/cm3 and Al radius = 143 pm.

Verify that Al is FCC. SOLUTION

4. Calculate number of Al per unit cell from mass of unit cell.

1 atom = 4.480 x 10-23 g, so

...more in lab and problem set

Mass 1 Al atom = 26.98 gmol

• 1 mol6.022 x 1023 atoms

1.79 x 10-22 gunit cell

• 1 atom4.480 x 10-23 g

= 3.99 Al atoms/unit cell

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Types of Crystalline

Solids

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Covalent-Network andMolecular Solids

Diamonds are an example of a covalent-network solid, in which atoms are covalently bonded to each other.

Covalent-network solids tend to be hard and have high melting points.

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Covalent-Network andMolecular Solids

Graphite is an example of a molecular solid, in which atoms are held together with intermolecular forces.

Molecular solids tend to be softer and have lower melting points.

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Metallic ForcesMetals have loosely held outer electrons - a sea of electrons over metal "ions"

Easy to melt - sea of electrons flows over solid or liquid

Conductors of electricity and heat - moving electrons in the "sea"

Malleable - not rigid like ionics, able to reform with sea still surrounding atoms

Difficult to boil - need to remove atom from "electron sea", high temperatures

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Ionic Solids

Na+ - Cl- in salt - the Ionic force

These are the strongest forces.

Lead to solids with high melting temperatures.

NaCl, mp = 800 oC MgO, mp = 2800 oC



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Comparing NaCl and CsCl

Even though their formulas have one cation and one anion, the lattices of CsCl and NaCl are different.

The different lattices arise from the fact that a Cs+ ion is much larger than a Na+ ion.

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End of Chapter 10

See: • Chapter Ten Study Guide • Chapter Ten Concept Guide



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